refactor: MaximumSumOfDistinctSubarraysWithLengthK (#5433)

* refactor: MaximumSumOfDistinctSubarraysWithLengthK

* checkstyle: fix formatting

* checkstyle: fix formatting

* checkstyle: fix formatting

---------

Co-authored-by: alxkm <[email protected]>

Author: alxkm

src/main/java/com/thealgorithms/others/MaximumSumOfDistinctSubarraysWithLengthK.java

@@ -1,55 +1,53 @@
 package com.thealgorithms.others;
 
 import java.util.HashSet;
+import java.util.Set;
 
-/*
-References: https://en.wikipedia.org/wiki/Streaming_algorithm
-* In this model, the function of interest is computing over a fixed-size window in the stream. As the stream progresses,
-* items from the end of the window are removed from consideration while new items from the stream take their place.
-* @author Swarga-codes (https://github.com/Swarga-codes)
-*/
+/**
+ * References: https://en.wikipedia.org/wiki/Streaming_algorithm
+ *
+ * This model involves computing the maximum sum of subarrays of a fixed size \( K \) from a stream of integers.
+ * As the stream progresses, elements from the end of the window are removed, and new elements from the stream are added.
+ *
+ * @author Swarga-codes (https://github.com/Swarga-codes)
+ */
 public final class MaximumSumOfDistinctSubarraysWithLengthK {
     private MaximumSumOfDistinctSubarraysWithLengthK() {
     }
-    /*
-     * Returns the maximum sum of subarray of size K consisting of distinct
-     * elements.
-     *
-     * @param k size of the subarray which should be considered from the given
-     * array.
+
+    /**
+     * Finds the maximum sum of a subarray of size K consisting of distinct elements.
      *
-     * @param nums is the array from which we would be finding the required
-     * subarray.
+     * @param k The size of the subarray.
+     * @param nums The array from which subarrays will be considered.
      *
-     * @return the maximum sum of distinct subarray of size K.
+     * @return The maximum sum of any distinct-element subarray of size K. If no such subarray exists, returns 0.
      */
     public static long maximumSubarraySum(int k, int... nums) {
         if (nums.length < k) {
             return 0;
         }
-        long max = 0; // this will store the max sum which will be our result
-        long s = 0; // this will store the sum of every k elements which can be used to compare with
-                    // max
-        HashSet<Integer> set = new HashSet<>(); // this can be used to store unique elements in our subarray
-        // Looping through k elements to get the sum of first k elements
+        long masSum = 0; // Variable to store the maximum sum of distinct subarrays
+        long currentSum = 0; // Variable to store the sum of the current subarray
+        Set<Integer> currentSet = new HashSet<>(); // Set to track distinct elements in the current subarray
+
+        // Initialize the first window
         for (int i = 0; i < k; i++) {
-            s += nums[i];
-            set.add(nums[i]);
+            currentSum += nums[i];
+            currentSet.add(nums[i]);
         }
-        // Checking if the first kth subarray contains unique elements or not if so then
-        // we assign that to max
-        if (set.size() == k) {
-            max = s;
+        // If the first window contains distinct elements, update maxSum
+        if (currentSet.size() == k) {
+            masSum = currentSum;
         }
-        // Looping through the rest of the array to find different subarrays and also
-        // utilising the sliding window algorithm to find the sum
-        // in O(n) time complexity
+        // Slide the window across the array
         for (int i = 1; i < nums.length - k + 1; i++) {
-            s = s - nums[i - 1];
-            s = s + nums[i + k - 1];
+            // Update the sum by removing the element that is sliding out and adding the new element
+            currentSum = currentSum - nums[i - 1];
+            currentSum = currentSum + nums[i + k - 1];
             int j = i;
             boolean flag = false; // flag value which says that the subarray contains distinct elements
-            while (j < i + k && set.size() < k) {
+            while (j < i + k && currentSet.size() < k) {
                 if (nums[i - 1] == nums[j]) {
                     flag = true;
                     break;
@@ -58,17 +56,14 @@ public static long maximumSubarraySum(int k, int... nums) {
                 }
             }
             if (!flag) {
-                set.remove(nums[i - 1]);
+                currentSet.remove(nums[i - 1]);
             }
-            set.add(nums[i + k - 1]);
-            // if the subarray contains distinct elements then we compare and update the max
-            // value
-            if (set.size() == k) {
-                if (max < s) {
-                    max = s;
-                }
+            currentSet.add(nums[i + k - 1]);
+            // If the current window has distinct elements, compare and possibly update maxSum
+            if (currentSet.size() == k && masSum < currentSum) {
+                masSum = currentSum;
             }
         }
-        return max; // the final maximum sum
+        return masSum; // the final maximum sum
     }
 }

src/test/java/com/thealgorithms/others/MaximumSumOfDistinctSubarraysWithLengthKTest.java

@@ -2,31 +2,21 @@
 
 import static org.junit.jupiter.api.Assertions.assertEquals;
 
-import org.junit.jupiter.api.Test;
+import java.util.stream.Stream;
+import org.junit.jupiter.params.ParameterizedTest;
+import org.junit.jupiter.params.provider.Arguments;
+import org.junit.jupiter.params.provider.MethodSource;
 
 public class MaximumSumOfDistinctSubarraysWithLengthKTest {
-    @Test
-    public void sampleTestCase1() {
-        assertEquals(15, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, 1, 5, 4, 2, 9, 9, 9));
-    }
-
-    @Test
-    public void sampleTestCase2() {
-        assertEquals(0, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, 4, 4, 4));
-    }
-
-    @Test
-    public void sampleTestCase3() {
-        assertEquals(12, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, 9, 9, 9, 1, 2, 3));
-    }
 
-    @Test
-    public void edgeCase1() {
-        assertEquals(0, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(0, 9, 9, 9));
+    @ParameterizedTest
+    @MethodSource("inputStream")
+    void testMaximumSubarraySum(int expected, int k, int[] arr) {
+        assertEquals(expected, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(k, arr));
     }
 
-    @Test
-    public void edgeCase2() {
-        assertEquals(0, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(5, 9, 9, 9));
+    private static Stream<Arguments> inputStream() {
+        return Stream.of(Arguments.of(15, 3, new int[] {1, 5, 4, 2, 9, 9, 9}), Arguments.of(0, 3, new int[] {4, 4, 4}), Arguments.of(12, 3, new int[] {9, 9, 9, 1, 2, 3}), Arguments.of(0, 0, new int[] {9, 9, 9}), Arguments.of(0, 5, new int[] {9, 9, 9}), Arguments.of(9, 1, new int[] {9, 2, 3, 7}),
+            Arguments.of(15, 5, new int[] {1, 2, 3, 4, 5}), Arguments.of(6, 3, new int[] {-1, 2, 3, 1, -2, 4}), Arguments.of(10, 1, new int[] {10}), Arguments.of(0, 2, new int[] {7, 7, 7, 7}), Arguments.of(0, 3, new int[] {}), Arguments.of(0, 10, new int[] {1, 2, 3}));
     }
 }